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Incident wave y= A sin (ax + bt+ pi/2) i...

Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by
`y= -1.6 sin ax sin bt + cA cos (bt + ax)`
where A, a, b and c are positive constants.
1. Amplitude of reflected wave is

A

0.6A

B

0.8A

C

0.4A

D

0.2A

Text Solution

Verified by Experts

The correct Answer is:
B
Reflected and incident rays are in the same
medium. Hence,
`I prop A^2`
`I_r has become 64% or 0.64 times of I_i`
`:. A_r = 0.8 A_i = 0.8 A ` .
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Knowledge Check

  • Incident wave y= A sin (ax + bt+ pi/2) is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by y= -1.6 sin ax sin bt + cA cos (bt + ax) where A, a, b and c are positive constants. 2. Value of c is

    A
    0.2
    B
    0.4
    C
    0.6
    D
    0.3

  • Incident wave y= A sin (ax + bt+ pi/2) is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by y= -1.6 sin ax sin bt + cA cos (bt + ax) where A, a, b and c are positive constants. 3. Position of second antinode is

    A
    `x = pi/(3a)`
    B
    `x = (3pi)/a`
    C
    `x = (3pi)/(2a)`
    D
    `x = (2pi)/(3a)`

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    A
    y = A sin (kx + `omega`t)
    B
    y = A cos (kx + `omega`t)
    C
    y = - A sin (kx - `omega`t)
    D
    y = - A sin (kx + `omega`t)

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